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D-Wave Systems Inc. has announced that it has built a quantum annealer with over 1000 qubits. Previously, D-Wave promised to deliver 1152 qubit systems in March 2015.
D-Wave's quantum computer runs a quantum annealing algorithm to find the lowest points, corresponding to optimal or near optimal solutions, in a virtual "energy landscape." Every additional qubit doubles the search space of the processor. At 1000 qubits, the new processor considers 21000 possibilities simultaneously, a search space which dwarfs the 2512 possibilities available to the 512-qubit D-Wave Two. In fact, the new search space contains far more possibilities than there are particles in the observable universe.
The new processors comprise over 128,000 Josephson junctions (tunnel junctions with superconducting electrodes) in a 6-metal layer planar process with 0.25µm features, believed to be the most complex superconductor integrated circuits ever built.
The press release goes on to explain that the new generation of D-Wave processors runs at a 40% colder temperature (closer to absolute zero than before), reduces noise levels, and allows for "new modes of use."
A blog post explains that specific delivered systems will vary in qubit count, but should have around 1152 qubits.
Original Submission
(Score: 2) by MichaelDavidCrawford on Tuesday June 23 2015, @06:31PM
i cant be bothered with wikipedia. how do these work?
my father used to use analog computers to calculate missile trajectories. at the time they were faster and cheaper than digital.
Yes I Have No Bananas. [gofundme.com]
(Score: 1, Informative) by Anonymous Coward on Tuesday June 23 2015, @07:20PM
A "true" quantum computer would use quantum states (qbits) which can act as a regular bit but also exhibits behavior rooted in quantum mechanics. In essence, they are physical representation of a bit which is also able to attain superposition just like regular quantum particles.
The fundamental idea of QCs is special logic gates [wikipedia.org], which in turn lets us create algorithms [wikipedia.org] that are impossible in traditional digital computers.
(Score: 4, Informative) by LoRdTAW on Tuesday June 23 2015, @07:34PM
An analog computer is programmable but not easily programmable like a microprocessor. Ever hear of an operational amplifier or op-amp? The operation part of the name means mathematic operations. They can be setup to add, subtract, multiply, divide, compare, invert, integrate, derivative etc. voltage and/or current become your variables that are fed into these circuits so you can build a circuit to solve an equation. As you adjust your input variables (voltages/currents) the various signals are manipulated and one or more outputs gives you a result. The downside is they aren't very programmable as you have to adjust component values or add/remove entire circuits to make a program. And before electronics, mechanical systems were built and it took hours to program them to perform a single calculation.
In ye-old days it made sense to solve simple problems like the PID loop of a generator or motor controller using analog computer circuits. That was when digital computers themselves were the size of rooms or refrigerators and probably used as much or more power than the device being controlled. Then as digital computers shrank they became more practical for use in control systems. Today? An arduino based on an 8-bit Atmel AVR microcontroller has built in counters, timers, analog-digital converters, serial ports and digital lines while running at 16MHz or more. That little chip could run a few dozen PID loops, take care of logic and remote digital communications for a dollar something per chip in small quantities. And it is completely programmable allowing on the fly adjustments which is something an analog computer can't practically do.
(Score: 0) by Anonymous Coward on Tuesday June 23 2015, @08:30PM
Beware of programmers that carry wire cutters in their pockets.
(Score: 5, Informative) by maxwell demon on Tuesday June 23 2015, @09:38PM
Quantum computers use the so-called superposition principle of quantum mechanics. Now what does that mean?
Imagine you've got several states that you can reliably distinguish. For example, consider a bit. A bit can have two clearly distinguishable states: It can be 0, or it can be 1. The qubit (quantum bit) has then also those states. Corresponding with the general usage in quantum mechanics, I'll in the following write states in the form |…〉. That is, our qubit has the states |0〉 and |1〉.
However, the superposition principle says that if you have those states, you can get more states by just doing linear combinations, that is, multiply them with numbers and add them. For example, the qubit can also be in the state 3|0〉+4|1〉 or in the state 7|0〉−|1〉 (now you see why the |...〉 notation is useful: Numbers inside and outside have completely different meanings). Indeed, the numbers can even be complex. Note however that states which differ only by a common non-zero factor are actually the same, that is, the state 3|0〉+4|1〉 is the same as the state 6|0〉+8|1〉 or the state ⅗|0〉 + ⅘|1〉 (this is quite similar to how the fractions 3/4 and 6/8 describe the same number). Note that the latter has a special property: |⅗|² + |⅘|² = 1. Such a representation is called normalized. (If you wonder about the absolute value bars: Those are needed for the case of complex numbers).
OK, so I've mathematically written the states down. But what do those states mean? Well, let's look at what happens if we ask whether the qubit is |0〉 or |1〉. Obviously, if the state is |0〉, we should find that it is |0〉, and if the state is |1〉, we should find that it is |1〉. But what if the state is ⅗|0〉 + ⅘|1〉? Well, quantum mechanics tells us that with probability |⅗|² we will find that it is in the state |0〉, and with probability |⅘|² we will find that it is in the state |1〉. It works this way of course only for normalized representations.
OK, so have we just written probabilities in a fancy way? At this point is looks that way, but is really isn't. A simple counterexample are the superposition states |0>+|1> and |0>−|1> (in normalized form they have an additional factor 1/sqrt(2) each). Quite obviously, for either of those states we have equal probability to find |0〉 or |1〉. However the states are clearly different, as you cannot get from one to another using a common factor. So how can we distinguish them?
Well, let's look what else we can do with a qubit. In particular, we are interested in reversible operations which furthermore don't change the normalization. For a classical bit, there are only two reversible operations: Either you do nothing, or you exchange 0 and 1. Of course you can do those operations also with a qubit. However you can do a lot more; and the power of a quantum computer is indeed due to that fact.
If we look at the description above with a mathematician's eye, we notice that what I described there is a (complex) vector space: You can add vectors and multiply them with a scalar. States are then represented by rays in that vector space. Now if you ask a mathematician what reversible operations don't change the normalization of vectors, he will tell you that those are the unitary operations (and the anti-unitary operations, but we will not consider those, because it turns out you cannot implement them physically). Unitary operations are simply rotations of the complex vector space.
Now you can see how we can distinguish the two vectors |0>+|1> and |0>−|1>; As one can easily check, those vectors are orthogonal to each other, and therefore we can just rotate them so that |0〉+|1〉 goes to (a multiple of) |0〉 and |0〉&minus|1〉 goes to (a multiple of) |1〉. Then we can just measure if we have |0〉 or |1〉, and then know for sure whether we had the state |0〉+|1〉 or |0〉−|1〉. Since we will need the states again further down, let's give them (or rather their normalized form) names: We define |+〉=(|0〉+|1〉)/sqrt(2) and |−〉=(|0〉−|1〉)/sqrt(2)
OK, so how can we use this for computation? Well, let's consider a simple problem: We have a function that takes one (classical) bit as argument and gives a one-bit result. Clearly, there exist four different such functions: Two that are constant, and in addition the identity function and the bit flip. Now we are given one of the functions, but are not told which one. And then we are asked to figure out whether it is a constant function.
With a classical computer there's only one way to answer this question: Compute the function for 0, compute it again for 1, and compare the results. If they are equal, the function is constant, otherwise it isn't. Clearly this needs two function evaluations. Can we do better with quantum computers?
Apart from the final measurement, all operations in a quantum computer are reversible. The constant functions are clearly not reversible, so we need to rewrite the functions to be reversible. This can be done by having *two* bits as input. The first bit represents the argument, and is not changed by the function, The second bit represents the output and is flipped if the function output is 1; so if it starts out as 0, it will contain the function input. Written that way, the function is clearly reversible; indeed, you can reverse it by simply applying it again.
OK, what does this mean in a quantum computer? Well, the distinguishable two-qubit states are |00〉, |01〉, |10〉 and |11〉, and for those states the function acts the very same as before. However remember the superposition principle: Every linear combination is also a valid state.
So let's start with the state |+−〉 = (|00〉−|01〉+|10〉&minus|11〉)/2. Now apply the function. Let's try all four possibilities separately:
So after applying any of the constant functions, the first qubit is in the state |+〉, while after applying one of the non-constant functions, the first qubit is in the state |−〉. As we have seen above, those two states can be perfectly distinguished, and therefore we indeed need only one application of the function to the superposition state to find out whether it is constant.
(Score: 2) by TheGratefulNet on Tuesday June 23 2015, @11:04PM
so, what happens if you ask the quantum computer to divide by 0?
"It is now safe to switch off your computer."
(Score: 0) by Anonymous Coward on Wednesday June 24 2015, @12:00AM
A quantum demon will be stuck in a superposition between being in your nose and coming out of it. The situation will resolve itself as soon as you observe it.
(Score: 2, Funny) by Wierd0n3 on Wednesday June 24 2015, @12:30AM
then McFly starts disappearing off the photo again
(Score: 0) by Anonymous Coward on Tuesday June 23 2015, @11:49PM
It's unfortunate that slashcode ate some of your forumals... I was barely following along, but it doesn't seem that hard if I studied it a bit.
(Score: 0) by Anonymous Coward on Wednesday June 24 2015, @01:03PM
I see the formulas quite fine. There's twice a "&minus" that obviously lacks a semicolon to produce − and a 0 in dollar signs proving LaTeX/MathJax habits, but otherwise I don't see any place where a formula seems to be missing or defaced. The formulas heavily use Unicode, though, thus maybe you are missing some Unicode fonts for a proper display.
Anyway, it's impossible that Slashcode ate anything since this site now runs on Rehash. ;-)